SOLUTIONS TO PROBLEMS ELEMENTARY LINEAR ...

1 SOLUTIONS TO PROBLEMSELEMENTARYLINEAR ALGEBRAK. R. MATTHEWSDEPARTMENT OF MATHEMATICSUNIVERSITY OF QUEENSLANDF irst Printing,1991 CONTENTSPROBLEMS (i) 0 0 02 4 0 R1 R2 2 4 00 0 0 R1 12R1 1 2 00 0 0 ;(ii) 0 1 31 2 4 R1 R2 1 2 40 1 3 R1 R1 2R2 1 0 20 1 3 ;(iii) 1 1 11 1 01 0 0 R2 R2 R1R3 R3 R1 1 1 00 0 10 1 1 R1 R1+R3R3 R3R2 R3 1 0 00 1 10 0 1 R2 R2+R3R3 R3 1 0 00 1 00 0 1 ;(iv) 2 0 00 0 0 4 0 0 R3 R3+ 2R1R1 12R1 1 0 00 0 00 0 0 .3. (a) 1 1 1 22 3 1 81 1 1 8 R2 R2 2R1R3 R3 R1 1 1 120 1 340 2 2 10 R1 R1 R2R3 R3+ 2R2 1 0 4 20 1 3 40 0 8 2 R3 18R3 1 0 4 20 1 3 40 0 114 R1 R1 4R3R2 R2+ 3R3 1 0 0 30 1 01940 0 114 .The augmented matrix has been converted to reduced row echelon formand we read off the unique solutionx= 3, y=194, z=14.

2 (b) 1 1 1 2 103 17 4 1 5 3 15 6 9 R2 R2 3R1R3 R3+ 5R1 1 1 1 2 100 4 10 2 290 8 20 4 59 R3 R3+ 2R2 1 1 1 2 100 4 10 2 290 0 0 01 .From the last matrix we see that the original system is (c) 3 1 7 02 1 4121 1 1 16 4 10 3 R1 R3 1 1 1 12 1 4123 1 7 06 4 10 3 R2 R2 2R1R3 R3 3R1R4 R4 6R1 1 1 1 10 1 2 320 2 4 30 2 4 3 R1 R1+R2R4 R4 R3R3 R3 2R2 1 0 3 120 1 2 320 0 0 00 0 0 0 .The augmented matrix has been converted to reduced row echelon formand we read off the complete solutionx= 12 3z, y= 32 2z, 2 1 3a3 1 5b 5 5 21c R2 R2 R1 2 1 3a1 2 8b a 5 5 21c R1 R2 1 2 8b a2 1 3a 5 5 21c R2 R2 2R1R3 R3+ 5R1 1 2 8b a0 5 19 2b+ 3a0 5 19 5b 5a+c R3 R3+R2R2 15R2 1 2 8b a0 1 1952b 3a50 00 3b 2a+c R1 R1 2R2 1 0 25(b+a)50 1 1952b 3a50 00 3b 2a+c.

3 From the last matrix we see that the original system is inconsistent if3b 2a+c6= 0. If 3b 2a+c= 0, the system is consistent and the solutionisx=(b+a)5+25z, y=(2b 3a)5+195z,wherezis 1 1 1t1t1 +t2 3 R2 R2 tR1R3 R3 (1 +t)R1 1 110 1 t00 1 t2 t R3 R3 R2 1 110 1 t00 0 2 t = 2. No 1 0 10 1 00 0 0 1 0 10 1 00 0 0 .We read off the unique solutionx= 1, y= Method 1. 3 1 1 11 3 1 11 1 3 11 1 1 3 R1 R1 R4R2 R2 R4R3 R3 R4 4 0 0 40 4 0 40 0 4 41 1 1 3 1 0 0 10 1 0 10 0 1 11 1 1 3 R4 R4 R3 R2 R1 1 0 0 10 1 0 10 0 1 10 0 0 0 .Hence the given homogeneous system has complete solutionx1=x4, x2=x4, x3=x4, 2. Write the system asx1+x2+x3+x4= 4x1x1+x2+x3+x4= 4x2x1+x2+x3+x4= 4x3x1+x2+x3+x4= it is immediate that any solution must satisfyx1=x2=x3= , ifx1, x2, x3, x4satisfyx1=x2=x3=x4, we get a 3 11 3 R1 R2 1 3 3 1 R2 R2 ( 3)R1 1 30 2+ 6 8 = 1: 2+ 6 86= 0.

4 That is ( 2)( 4)6= 0 or 6= 2,4. HereBisrow equivalent to 1 00 1 :R2 1 2+6 8R2 1 30 1 R1 R1 ( 3)R2 1 00 1 .Hence we get the trivial solutionx= 0, y= 2: = 2. ThenB= 1 10 0 and the solution isx=y, 3: = 4. ThenB= 1 10 0 and the solution isx= y, 3 1 1 15 1 1 1 R1 13R1 11313135 1 1 1 R2 R2 5R1 11313130 83 23 83 R2 38R2 11313130 1141 R1 R1 13R2 1 01400 1141 .Hence the solution of the associated homogeneous system isx1= 14x3, x2= 14x3 x4, 1 n1 11 1 n ..11 1 n R1 R1 RnR2 R2 1 Rn 1 Rn n0 n0 n ..1 1 1 n 1 0 10 1 ..1 1 1 n Rn Rn Rn 1 R1 1 0 10 1 ..0 0 0 .The last matrix is in reduced row echelon the homogeneous system with coefficient matrixAhas thesolutionx1=xn, x2=xn.

5 , xn 1=xn, , writing the system in the formx1+ +xn=nx1x1+ +xn= + +xn=nxnshows that any solution must satisfynx1=nx2= =nxn, sox1=x2= =xn. Conversely ifx1=xn, .. , xn 1=xn, we see thatx1, .. , xnis LetA= a bc d and assume thatad bc6= 1:a6= 0. a bc d R1 1aR1 1bac d R2 R2 cR1 1ba0ad bca R2 aad bcR2 1ba0 1 R1 R1 baR2 1 00 1 .Case 2:a= 0. Thenbc6= 0 and hencec6= 0bc d R1 R2 c d0b 1dc0 1 1 00 1 .So in both cases,Ahas reduced row echelon form equal to 1 00 1 .11. We simplify the augmented matrix of the system using row operations: 1 2 343 1524 1a2 14a+ 2 R2 R2 3R1R3 R3 4R1 1 2 340 7 14 100 7a2 2a 14 R3 R3 R2R2 17R2R1 R1 2R2 1 2 340 1 21070 0a2 16a 4 R1 R1 2R2 1 01870 1 21070 0a2 16a 4 .Denote the last matrix 1:a2 166= 0.

6 4. ThenR3 1a2 16R3R1 R1 R3R2 R2+ 2R3 1 0 08a+257(a+4)0 1 010a+547(a+4)0 0 11a+4 and we get the unique solutionx=8a+ 257(a+ 4), y=10a+ 547(a+ 4), z=1a+ 2:a= 4. ThenB= 1 0 1870 1 21070 0 0 8 , so our system is 3:a= 4. ThenB= 1 0 1870 1 21070 0 0 0 . We read off that the system isconsistent, with complete solutionx=87 z, y=107+ 2z, We reduce the augmented array of the system to reduced row echelonform: 1 0 1 0 10 1 0 1 11 1 1 1 00 0 1 1 0 R3 R3+R1 1 0 1 0 10 1 0 1 10 1 0 1 10 0 1 1 0 R3 R3+R2 1 0 1 0 10 1 0 1 10 0 0 0 00 0 1 1 0 R1 R1+R4R3 R4 1 0 0 1 10 1 0 1 10 0 1 1 00 0 0 0 0 .The last matrix is in reduced row echelon form and we read off the solutionof the corresponding homogeneous system:x1= x4 x5=x4+x5x2= x4 x5=x4+x5x3= x4=x4,6wherex4andx5are arbitrary elements ofZ2.

7 Hence there are four SOLUTIONS :x1x2x3x4x50 0 0 0 01 1 0 0 11 1 1 1 00 0 1 1 (a) We reduce the augmented matrix to reduced row echelon form: 2 1 3 44 1 4 13 1 2 0 R1 3R1 1 3 4 24 1 4 13 1 2 0 R2 R2+R1R3 R3+ 2R1 1 3 4 20 4 3 30 2 0 4 R2 4R2 1 3 4 20 1 2 20 2 0 4 R1 R1+ 2R2R3 R3+ 3R2 1 0 3 10 1 2 20 0 1 0 R1 R1+ 2R3R2 R2+ 3R3 1 0 0 10 1 0 20 0 1 0 .Consequently the system has the unique solutionx= 1, y= 2, z= 0.(b) Again we reduce the augmented matrix to reduced row echelon form: 2 1 3 44 1 4 11 1 0 3 R1 R3 1 1 0 34 1 4 12 1 3 4 R2 R2+R1R3 R3+ 3R1 1 1 0 30 2 4 40 4 3 3 R2 3R2 1 1 0 30 1 2 20 4 3 3 R1 R1+ 4R2R3 R3+R2 1 0 3 10 1 2 20 0 0 0.

8 We read off the complete solutionx= 1 3z= 1 + 2zy= 2 2z= 2 + 3z,wherezis an arbitrary element Suppose that ( 1, .. , n) and ( 1, .. , n) are SOLUTIONS of the systemof LINEAR equationsnXj=1aijxj=bi,1 i j=biandnXj=1aij j=bifor 1 i i= (1 t) i+t ifor 1 i m. Then ( 1, .. , n) is a solution ofthe given system. FornXj=1aij j=nXj=1aij{(1 t) j+t j}=nXj=1aij(1 t) j+nXj=1aijt j= (1 t)bi+tbi= Suppose that ( 1, .. , n) is a solution of the system of LINEAR equationsnXj=1aijxj=bi,1 i m.(1)Then the system can be rewritten asnXj=1aijxj=nXj=1aij j,1 i m,or equivalentlynXj=1aij(xj j) = 0,1 i we havenXj=1aijyj= 0,1 i j=yj. Hencexj= j+yj,1 j n, where (y1, .. , yn) isa solution of the associated homogeneous system. Conversely if (y1, .. , yn)8is a solution of the associated homogeneous system andxj= j+yj,1 j n, then reversing the argument shows that (x1.)

9 , xn) is a solution ofthe system 1 .16. We simplify the augmented matrix using row operations, working to-wards row echelon form: 1 1 1 1 1a1 1 1b3 2 0a1 +a R2 R2 aR1R3 R3 3R1 1 1 1110 1 a1 +a1 a b a0 13a 3a 2 R2 R3R2 R2 1 1 1110 1 3 3 a2 a0 1 a1 +a1 a b a R3 R3+ (a 1)R2 1 1 1110 1 33 a2 a0 0 4 2a(1 a)(a 2) a2+ 2a+b 2 = 1:a6= 2. Then 4 2a6= 0 andB 1 1 1 110 1 3 3 a2 a0 0 1a 12 a2+2a+b 24 2a .Hence we can solve forx, yandzin terms of the arbitrary 2:a= 2. ThenB= 1 1 1 1 10 1 3 1 00 0 0 0b 2 .Hence there is no solution ifb6= 2. However ifb= 2, thenB= 1 1 1 1 10 1 3 1 00 0 0 0 0 1 0 2 0 10 1 3 1 00 0 0 0 0 and we get the solutionx= 1 2z, y= 3z w, wherewis (a) We first prove that 1 + 1 + 1 + 1 = 0.

10 Observe that the elements1 + 0,1 + 1,1 +a,1 +b9are distinct elements ofFby virtue of thecancellation law for addition. Forthis law states that 1+x= 1+y x=yand hencex6=y 1+x6= 1+ the above four elements are just the elements 0,1, a, bin someorder. Consequently(1 + 0) + (1 + 1) + (1 +a) + (1 +b) = 0 + 1 +a+b(1 + 1 + 1 + 1) + (0 + 1 +a+b) = 0 + (0 + 1 +a+b),so 1 + 1 + 1 + 1 = 0 after 1 + 1 + 1 + 1 = (1 + 1)(1 + 1), so we havex2= 0, wherex= 1 + 0. Thena+a=a(1 + 1) =a 0 = +b= 1. Fora+bmust be one of 0,1, a, b. Clearly we can thavea+b=aorb; also ifa+b= 0, thena+b=a+aand henceb=a;hencea+b= 1. Thena+ 1 =a+ (a+b) = (a+a) +b= 0 +b= + 1 =a. Consequently the addition table forFis+ 0 1 a now find the multiplication table. First,abmust be one of 1, a, b;however we can t haveab=aorb, so this leavesab= Fora2must be one of 1, a, b; howevera2=a a= 0 ora= 1; alsoa2= 1 a2 1 = 0 (a 1)(a+ 1) = 0 (a 1)2= 0 a= 1;hencea2=b.